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Problem : For $g$ $\in$ $\mathbb{Z}$ , let $g' \in \mathbb{Z_{37}}$ denotes the residue class of $g$ mod $37$. Consider the group $U_{37}$ = {$g' \in \mathbb{Z_{37}} :1≤g≤37 \text{ with}, gcd(g, 37) = 1$} with respect to multiplication mod $37$. Then which one of the following is FALSE?

(A) The set {$g' \in U_{37} ∶ g'=g'^{-1}$} contains exactly 2 elements.

(B) The order of the element $10'$ in $U_{37}$ is $36$.

My Try: For option A , in a group of even order there are odd number of elements of order $2$ ,along with $e$ "identity" there will be even number of elements satisfying option A but how to confirm that there will only one element of order $2$ ?

For option B, I don't think that operating $10'$ in $U_{37}$, 36 times is a good approach to confirm that it is true or false . Is there any formula to calculate the order of elements in $U(n)$ or any good approach to do it ?

Gajjze
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1 Answers1

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For (A), it is worth noting that $U_{37}$ is cyclic and has order $\varphi(37)=36$. So if you have an element of order $2$, then it is necessarily unique.

For (B), the question is essentially asking if $10'$ is the residue class mod $37$ that generates this cyclic group. The prime factorization of $36$ is $2^23^2$ and if $10'$ has order less than $36$, then it must divide $36$.

The possible divisors one gets are therefore $1,2,3,4,9,12,18,36$. From here, it isn't too hard to check by hand and if you go in order from lowest degrees. For example, $10^{12}\equiv 1\bmod 37$ is the identity. The easiest way to see this is to compute $10^3\equiv 1\bmod 37$.

Shrugs
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